Question

Having landed on a newly discovered planet,

an astronaut sets up a simple pendulum of
length 1.04 m and finds that it makes 475
complete oscillations in 1320 s. The amplitude of the oscillations is very small compared
to the pendulum’s length.
What is the gravitational acceleration on
the surface of this planet?
Answer in units of m/s^2.

Answer 1

17.215 m/s^2

Step-by-step explanation:

L = 0.825 m

It completes 397 oscillations in 546 s.

Time period is defined as the time taken to complete one oscillation.

So, Time period, T = 546 / 397 = 1.375 s

Let g be the gravitational acceleration at that planet.

Use the formula for the time period

g = 17.215 m/s^2